Re: Perimeter of triangle

Get AM by Pythagoras (=8)
AngleACM=angleB (both=90-A)
sinB=6/CB But from triangle ACM sinACM=8/10 So sinB=8/10 So 8/10=6/CB So CB=15/2
In triangle BCM cosB=BM/CB So 6/10=BM/15/2 BM=9/2
Perimeter=AC+CB+BM+MA=10+7.5+4.5+8=30